One side of a triangle is 5cm and the other side is 10 cm and its perimter is P cm, where P is an integer. The least and the greatest possible values of P are respectively:

Options

A. 20 and 28
B. 21 and 29
C. 22 and 27
D. 19 and 29

Answer

To find the least and greatest possible values of the perimeter \( P \) of a triangle with sides measuring 5 cm and 10 cm, we will denote the third side as \( X \) cm. The perimeter \( P \) can be expressed as:

\[ P = 5 + 10 + X = 15 + X \]

### Step 1: Determine the minimum value of \( X \)

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, we have:

1. \( 5 + X > 10 \)
2. \( 10 + X > 5 \)
3. \( 5 + 10 > X \)

From the first inequality:

\[ X > 10 - 5 \implies X > 5 \]

From the second inequality:

\[ X > 5 - 10 \implies X > -5 \quad (\text{This is always true since } X > 5) \]

From the third inequality:

\[ X < 15 \]

Thus, we have:

\[ 5 < X < 15 \]

Since \( X \) must be an integer, the smallest integer value for \( X \) is 6.

### Step 2: Calculate the minimum value of \( P \)

Substituting \( X = 6 \) into the perimeter equation:

\[ P = 15 + 6 = 21 \]

### Step 3: Determine the maximum value of \( X \)

From the inequalities derived earlier, we know:

\[ X < 15 \]

The largest integer value for \( X \) is 14.

### Step 4: Calculate the maximum value of \( P \)

Substituting \( X = 14 \) into the perimeter equation:

\[ P = 15 + 14 = 29 \]

### Conclusion

The least and greatest possible values of \( P \) are:

\[ \text{Least value of } P = 21 \quad \text{and} \quad \text{Greatest value of } P = 29 \]